Rotational motion down an incline (ring/hoop)

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SUMMARY

The discussion focuses on the physics of a 24 kg metal ring with a 24 cm diameter rolling down a 30-degree incline from a height of 3.4 m. The conservation of energy principle is applied to determine the linear speed at the bottom of the ramp, leading to the conclusion that the speed is derived from the equation V = √(2gh) for a frictionless scenario. The moment of inertia, I = mr², is used to calculate the linear speed under the assumption of rolling without slipping, yielding a different speed calculation. The average linear acceleration of the ring down the incline is identified as gsin(θ), where θ is the incline angle.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with moment of inertia calculations (I = mr²)
  • Knowledge of angular velocity and its relationship to linear speed
  • Basic understanding of kinematics, specifically acceleration equations
NEXT STEPS
  • Study the derivation of the conservation of energy equation in rotational motion
  • Learn about the relationship between linear and angular acceleration in rolling objects
  • Explore the effects of friction on rolling motion and energy conservation
  • Investigate the kinematic equations linking initial speed, final speed, distance, and acceleration
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Students studying physics, particularly those focusing on rotational motion and energy conservation, as well as educators looking for examples of practical applications of these concepts.

floraha
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Homework Statement


A 24kg metal ring with 24cm diameter rolls without slipping down a 30 degree incline from a height of 3.4 m.

1) According to the law of conservation of energy what should be the linear speed of the ring at the bottom of the ramp
2) if the ring has a moment of inertia of I=mr^2 what will its linear speed be at the base of the incline?

3) what is the avg linear acc of this ring down the incline?

Homework Equations

3. The Attempt at a Solution [/B]
For part 1, I got the math down to the square root of 2gh, because I assumed they were just asking for Vfrictionless. I did this through the equation PE=KE(translational) + KE (rotational) from which I got mgh = 1/2mv^2+1/2iw^2 where i=moment of inertia and w=angular velocity. In this case (i) would just be zero, getting me 2gh

For part 2 I did the same thing, except I substituted the moment of inertia in for (i) getting the math down to the square root of gh.

I had the most difficult time with part three. I wasn't sure where to start so I just worked out a(tangental) = r * angular acceleration. My teacher told me the answer is gsin(x) but I have no idea how he got there.
 
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floraha said:
rolls without stopping
Are you sure that is the correct wording and it wasn't supposed to say, "rolls without slipping". Additionally, because it says "rolls", that suggests that it is not frictionless. If it was frictionless, it would not rotate at all; it would just slide.

Edit: Sorry I forgot to welcome you to Physics Forums. Welcome!
Edit2: Hmm. I think I can see your position that part 1 maybe is supposed to assume frictionless due to the wording of part 2. However, for part 1 the ring had the same moment of inertia as specified in part 2.
 
@TomHart Thank you for replying! You were right about the wording, it was actually "slipping." Also, for part 3 it autocorrected to arc when it should have been acc (as in acceleration).
 
floraha said:
what is the avg linear acc of this ring down the incline?
You know the initial speed, the final speed and the distance (which is...?). What equation do you know linking those three to acceleration?
 

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